Determine where $f(x)$ intersects the $x$ -axis. $f(x) = (x + 3)^2 - 64$
Solution: The function intersects the $x$ -axis where $f(x) = 0$ , so solve the equation: $ (x + 3)^2 - 64 = 0$ Add $64$ to both sides so we can start isolating $x$ on the left: $ (x + 3)^2 = 64$ Take the square root of both sides to get rid of the exponent. $ \sqrt{(x + 3)^2} = \pm \sqrt{64}$ Be sure to consider both positive and negative $8$ , since squaring either one results in $64$ $ x + 3 = \pm 8$ Subtract $3$ from both sides to isolate $x$ on the left: $ x = -3 \pm 8$ Add and subtract $8$ to find the two possible solutions: $ x = 5 \text{or} x = -11$